Matrices Test

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Matrices Test - 45

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Question 1
1 / -0

If $$O\left( A \right) =2\times 3,$$ $$O\left( B \right) =3\times 2$$ and $$O\left( C \right) =3\times 3$$, which one of the following is not defined?

A
$$CB+{ A }^{ ' }$$

B
$$BAC$$

C
$$C{ \left( A+{ B }^{ ' } \right) }^{ ' }$$

D
$$C\left( A+{ B }^{ ' } \right) $$

Solution

Given that, $$O\left( A \right) =2\times 3,$$ $$O\left( B \right) =3\times 2$$ and $$O\left( C \right) =3\times 3$$
$$\Rightarrow O\left( { A }^{ ' } \right) =3\times 2, O\left( { B }^{ ' } \right) =2\times 3$$
(a) $$CB+{ A }^{ ' }$$
Now, Order of $$CB=$$ (Order of $$C$$) (Order of $$B$$)
$$=$$ (Order of $$C$$ is $$3\times 3$$) (Order of $$B$$ is $$3\times 2$$)
$$=$$ Order of $$CB$$ is $$3\times 2$$
Since, $$O\left( { A }^{ ' } \right) =3\times 2$$
Therefore, matrix $$CB+{ A }^{ ' }$$ can be determined.
(b) $$O\left( BA \right) =3\times 3$$ and $$O\left( C \right) =3\times 3$$
Therefore, matrix $$BAC$$ can be determined.
(c) $$C{ \left( A+{ B }^{ ' } \right) }^{ ' }$$
$$O\left( A+{ B }^{ ' } \right) =2\times 3$$
$$\Rightarrow O{ \left( A+{ B }^{ ' } \right) }^{ ' }=3\times 2$$
and $$O\left( C \right) =3\times 3$$
Therefore, matrix $$C{ \left( A+{ B }^{ ' } \right) }^{ ' }$$ can be determined.
(d) $$C\left( A+{ B }^{ ' } \right) $$
$$O\left( A+{ B }^{ ' } \right) =2\times 3$$
and $$O\left( C \right) =3\times 3$$
$$\therefore $$ Matrix $$C\left( A+{ B }^{ ' } \right) $$ cannot be determined
Hence, option D is correct.
Question 2
1 / -0

If $$A=
\begin{bmatrix}
2 & -1 \\
-1 & 2
\end{bmatrix}$$ and $$I$$ is the unit matrix of order $$2$$, then $$A^2$$equals

A
$$4A-3I$$

B
$$3A-AI$$

C
$$A-I$$

D
$$A+I$$

Solution

$$A=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\\ { A }^{ 2 }=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\times \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\\ \quad =\begin{bmatrix} 4+1 & -2-2 \\ -2-2 & 1+4 \end{bmatrix}\\ \quad =\begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}$$
Answer will be A option because
$$4A-3I=\begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\\ \quad \quad \quad \ =\begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}$$
Question 3
1 / -0

Let $$A=\begin{bmatrix} 3 & 1\\ -1 & 2\end{bmatrix}$$, then

A
$$A^2+7A-5I=0$$

B
$$A^2-7A+5I=0$$

C
$$A^2+5A-7I=0$$

D
$$A^2-5A+7I=0$$

Solution

$${ A }^{ 2 }=\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}=\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$$
$${ A }= \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$$
$${ I }= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
We find (d) satisfy the condition as
$${ A }^{ 2 }-5A-7I=0$$
$$\quad \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}-\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}+\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}=0$$
Thus,$${ A }^{ 2 }-5A-7I=0$$ is correct
Question 4
1 / -0

Let $$A=\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$$ and $$10B=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$$. If $$B$$ is the inverse of matrix $$A$$, then $$\alpha $$ is

A
$$-2$$

B
$$1$$

C
$$2$$

D
$$5$$

Solution

Since, $$B$$ is the inverse of matrix $$A$$.
So, $$10{ A }^{ -1 }=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$$
$$\Rightarrow 10{ A }^{ -1 }\cdot A=\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$$
$$\Rightarrow 10I = \begin{bmatrix} 4+4+2 & -4+2+2 & 4-6+2 \\ -5+0+\alpha & 5+0+\alpha & -5+0+\alpha \\ 1-4+3 & -1-2+3 & 1+6+3 \end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix}=\begin{bmatrix} 10 & 0 & 0 \\ -5+\alpha & 5+\alpha & -5+\alpha \\ 0 & 0 & 10 \end{bmatrix}$$
$$\Rightarrow -5+\alpha =0$$
$$\Rightarrow \alpha =5$$
Question 5
1 / -0

$$If\quad A=\begin{bmatrix} ab & { b }^{ 2 } \\ -{ a }^{ 2 } & -ab \end{bmatrix},then\quad { A }^{ 2 }\quad is\quad equal\quad to$$

A
$$O$$

B
$$I$$

C
$$-I$$

D
$$None\quad of\quad these$$

Solution
Question 6
1 / -0

If $$\left[ \begin{matrix} x & 4 & -1 \end{matrix} \right] \left[ \begin{matrix} 2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 4 \end{matrix} \right] \left[ \begin{matrix} x \\ 4 \\ -1 \end{matrix} \right] =0,$$ then $$x=$$

A
$$-1+\sqrt { 6 } $$

B
$$8\pm \sqrt { 5 } $$

C
$$-2\pm \sqrt { 10 } $$

D
$$3\pm \sqrt { 6 } $$
Question 7
1 / -0

The value of $$x$$, so that $$\left[ 1 \quad x \quad 1 \right] \begin{bmatrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$$ is/are

A
$$\pm\ 2$$

B
$$0$$

C
$$\dfrac { -7\pm \sqrt { 35 } }{ 2 }$$

D
$$\dfrac { -9\pm \sqrt { 53 } }{ 2 }$$

Solution
Question 8
1 / -0

If the matrix $$\begin{bmatrix} 0 & 2\beta & \Upsilon \\ \alpha & \beta & -\Upsilon \\ \alpha & -\beta & \Upsilon \end{bmatrix}$$is orthogonal, then

A
$$\alpha = \pm\dfrac{1}{\sqrt{2}}$$

B
$$\beta = \pm\dfrac{1}{\sqrt{6}}$$

C
$$\gamma = \pm\dfrac{1}{\sqrt{3}}$$

D
all of these

Solution

$$\begin{bmatrix} 0 & 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{bmatrix}$$
for orthogonal matrix we have
$$A.A^{T}=I$$
$$\begin{bmatrix} 0 & 2\beta & \gamma \\ \alpha & \beta & -\gamma \\ \alpha & -\beta & \gamma \end{bmatrix}\begin{bmatrix} 0 & \alpha & \alpha \\ 2\beta & \beta & -\beta \\ \gamma & -\gamma & \gamma \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$\begin{bmatrix} 0+4{ \beta }^{ 2 }+{ \gamma }^{ 2 } & 0+2{ \beta }^{ 2 }-{ \gamma }^{ 2 } & 0-2{ \beta }^{ 2 }+{ \gamma }^{ 2 } \\ 0+2{ \beta }^{ 2 }-{ \gamma }^{ 2 } & { \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 } & { \alpha }^{ 2 }-{ \beta }^{ 2 }-{ \gamma }^{ 2 } \\ 0-2{ \beta }^{ 2 }+{ \gamma }^{ 2 } & { \alpha }^{ 2 }-{ \beta }^{ 2 }-{ \gamma }^{ 2 } & { \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 } \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$4\beta^{2}+\gamma^{2}=1, 2\beta^{2}-\gamma^{2}=0$$,
$$4\left(\dfrac{\gamma^{2}}{2}\right)+\gamma^{2}=1$$ $$\beta^{2}=\dfrac{r^{2}}{2}$$
$$r^{2}[3]=1$$
$$r=\pm \dfrac{1}{\sqrt{3}}$$
$$2\beta^{2}-\gamma^{2}=0, \alpha^{2}+\beta^{2}+\gamma^{2}+\gamma^{2}=1, \alpha^{2}-\beta^{2}-\gamma^{2}=0$$
$$\beta^{2}=\dfrac{\gamma^{2}}{2}, \alpha^{2}+\dfrac{\gamma^{2}}{2}+\dfrac{\gamma^{2}}{1}=1$$
$$\alpha^{2}+\dfrac{3\gamma^{2}}{2}=1$$
$$\beta^{2}=\dfrac{1}{6}\alpha^{2}+\dfrac{3}{2}\times \dfrac{1}{3}=1$$
$$\beta=\pm \dfrac{1}{\sqrt{6}}$$ $$\alpha=\pm \dfrac{1}{\sqrt{2}}$$
Question 9
1 / -0

If $$A=\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$$, then : $$A^{-1}$$=

A
$$A$$

B
$$A^{2}$$

C
$$A^{3}$$

D
$$A^{4}$$

Solution
Question 10
1 / -0

If $$A=\left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] $$, then $$A^{2}-4\ A$$ is equal to

A
$$2\ I_3$$

B
$$3\ I_3$$

C
$$4\ I_3$$

D
$$5\ I_3$$

Solution

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Re-Attempt Weekly Quiz Competition

Matrices Test - 45

Result

Matrices Test - 45

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Rank

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SHARING IS CARING

Question 1

$$CB+{ A }^{ ' }$$

$$BAC$$

$$C{ \left( A+{ B }^{ ' } \right) }^{ ' }$$

$$C\left( A+{ B }^{ ' } \right) $$

Solution

Question 2

$$4A-3I$$

$$3A-AI$$

$$A-I$$

$$A+I$$

Solution

Question 3

$$A^2+7A-5I=0$$

$$A^2-7A+5I=0$$

$$A^2+5A-7I=0$$

$$A^2-5A+7I=0$$

Solution

Question 4

$$-2$$

$$1$$

$$2$$

$$5$$

Solution

Question 5

$$O$$

$$I$$

$$-I$$

$$None\quad of\quad these$$

Solution

Question 6

$$-1+\sqrt { 6 } $$

$$8\pm \sqrt { 5 } $$

$$-2\pm \sqrt { 10 } $$

$$3\pm \sqrt { 6 } $$

Question 7

$$\pm\ 2$$

$$0$$

$$\dfrac { -7\pm \sqrt { 35 } }{ 2 }$$

$$\dfrac { -9\pm \sqrt { 53 } }{ 2 }$$

Solution

Question 8

$$\alpha = \pm\dfrac{1}{\sqrt{2}}$$

$$\beta = \pm\dfrac{1}{\sqrt{6}}$$

$$\gamma = \pm\dfrac{1}{\sqrt{3}}$$

all of these

Solution

Question 9

$$A$$

$$A^{2}$$

$$A^{3}$$

$$A^{4}$$

Solution

Question 10

$$2\ I_3$$

$$3\ I_3$$

$$4\ I_3$$

$$5\ I_3$$

Solution

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